[John wrote](https://forum.azimuthproject.org/discussion/comment/19191/#Comment_19191):
> I'm not sure what "do this again" means. I would love to get [hyperoperations](https://en.wikipedia.org/wiki/Hyperoperation#Examples) this way. But there are some puzzling features.

Hmm, you're quite right! I saw the start of a pattern and charged ahead rather blindly. What I wanted to do was iteratively consider the sequence of categories \$$\mathcal{X}\_n\$$ with morphisms all functors \$$F\_n : \mathbf{N} \to \mathcal{X}\_{n-1}\$$. But once you get \$$\mathcal{X}\_2 = \mathbf{Exp}\$$, the category with two objects consisting of functors \$$\mathbf{N} \to \mathbf{Mult}\$$, you no longer have any useful cycles, so you can no longer iterate your morphisms, and everything locks up. (I think \$$\mathcal{X}\_3 = \mathbf{Unit}\$$, if we forced onwards with functors \$$\mathbf{N} \to \mathbf{Exp}\$$, which is a good sign we've gone as far as we can.)

> **Puzzle.** What is the category with \$$\mathbf{M}\$$ as its only object and functors from \$$\mathbf{M}\$$ to itself as morphisms?

Firstly, thanks to the prime decomposition theorem, we can treat \$$\mathbf{M}\$$ as generated by an infinite set of elements: the prime numbers. So this gives us something to hang on to when considering the action of our functors. It isn't a finite presentation, but we should still have as a theorem that every arrow can be factored into a finite set of primes.

Therefore, a functor \$$F : \mathbf{M} \to \mathbf{M}\$$ is determined by its action on the prime numbers. One interesting thing to note is that two functors give isomorphic images if their action on the primes is different only by a permutation of the primes. For instance, if \$$F\$$ sends \$$2\$$ to \$$42\$$ and \$$3\$$ to \$$101\$$, \$$F'\$$ sends \$$3\$$ to \$$42\$$ and \$$2\$$ to \$$101\$$, and both fix all other primes, then they only differ by the permutation \$$(42\ 101)\$$. So it seems that permutations on the set of prime numbers give some kind of equivalence between functors.

So it suffices to consider functors based on how much they flatten the set of primes. If \$$\lvert F(\mathbf{Prime}) \rvert = \lvert \mathbf{Prime} \rvert\$$, we get a category equivalent to what we started with. If, however, \$$\lvert F(\mathbf{Prime}) \rvert = k\$$ for finite \$$k\$$, then the image category has only a finite set of generators.

This seems like a kind of projection, or almost a kind of division, but it's sufficiently strange that I'm not sure what to do with it.